The book says it follows from this theorem: Suppose $f$ is integrable on $X$, $\{X_n\}_{n=1}^\infty$, a disjoint countable colletction of measurable sets whose union is $X$. Then $$\int _X f~d\mu = \sum _{n=1}^\infty \int_{X_n} f~d\mu.$$ I however can't see it, and I ask for your help.

2 Answers
2

Hint for 1.: If $(X_n)_{n\geqslant1}$ is ascending with union $X$ and if every $X_n$ is measurable, then $Y_1=X_1$ and $Y_{n+1}=X_{n+1}\setminus X_{n}$ for every $n\geqslant1$ defines a sequence $(Y_n)_{n\geqslant1}$ of disjoint measurable subsets with union $X$.

Hint for 2.: If $(X_n)_{n\geqslant1}$ is descending with intersection $X$ and if every $X_n$ is measurable, then $Y_{n}=X_{1}\setminus X_{n+1}$ for every $n\geqslant1$ defines an ascending sequence $(Y_n)_{n\geqslant1}$ of measurable subsets with union $X_{1}\setminus X$. Furthermore, $f$ is integrable on $X_1$ hence $\int\limits_{X_1\setminus X}f\mathrm d\mu=\int\limits_{X_1}f\mathrm d\mu-\int\limits_{X}f\mathrm d\mu$ and, for every $n\geqslant1$, $\int\limits_{X_1\setminus X_{n+1}}f\mathrm d\mu=\int\limits_{X_1}f\mathrm d\mu-\int\limits_{X_{n+1}}f\mathrm d\mu$.

We just "disjoint-ify" the ascending / descending collections of sets:

If $X_1\subseteq X_2\subseteq \cdots \subseteq X$ is an ascending (countable) sequence of sets with
$$\bigcup_{n=1}^\infty X_n=X,$$
then let $Y_1=X_1$ and let $Y_n=X_n\setminus X_{n-1}$ for all $n>1$. Thus, for any $n$ we have that
$$X_n=\bigcup_{i=1}^n Y_i,$$
and this is a disjoint union, so that
$$\int_{X_n}f\,d\mu=\sum_{i=1}^n\int_{Y_{\,i}}f\,d\mu.$$
Because we now have a countable disjoint collection of sets $\{Y_i\}$ that have the property$$\bigcup_{n=1}^\infty Y_i=X,$$
by the cited theorem we know that
$$\int_X f\,d\mu=\sum_{i=1}^\infty\int_{Y_i}f\,d\mu=\lim_{n\to\infty}\sum_{i=1}^n\int_{Y_i}f\,d\mu=\lim_{n\to\infty}\int_{X_n}f\,d\mu.$$